Let $x_{1},x_{2},\cdots\cdot x_{n}$ be $\mathrm{n}$ observations, and let $\bar{x}$ be their arithmetic mean and $\sigma^{2}$ be their variance. Statement 1: Variance of $2x_{1},2x_{2},\ldots,2x_{n}$ is $4\sigma^{2}$. Statement 2: Arithmetic mean of $2x_{1},2x_{2},\ldots,2x_{n}$ is $4\bar{x}$.

Using the properties of variance and mean under scaling, if each observation is multiplied by a constant k, the new variance becomes k squared times the original variance. For the observations 2 times x, the variance is 2 squared times sigma squared, which equals 4 sigma squared, making statement 1 true. The arithmetic mean of the new observations becomes k times the original mean, meaning the mean of 2 times x is 2 times x bar, not 4 times x bar, making statement 2 false. Therefore, statement 1 is true and statement 2 is false.

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